Tìm nghiệm ∈ ( -5 $\pi$ /2; 3 $\pi$ ) khi a) sinx=1/2 b) sin(x+ $\pi$ /4)=0 27/07/2021 Bởi Hadley Tìm nghiệm ∈ ( -5 $\pi$ /2; 3 $\pi$ ) khi a) sinx=1/2 b) sin(x+ $\pi$ /4)=0
Đáp án: $\begin{array}{l}a)\sin x = \dfrac{1}{2}\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\Do: – \dfrac{{5\pi }}{2} < x < 3\pi \\ \Rightarrow x \in \left\{ {\dfrac{{ – 11\pi }}{6};\dfrac{\pi }{6};\dfrac{{13\pi }}{6};\dfrac{{ – 7\pi }}{6};\dfrac{{5\pi }}{6};\dfrac{{17\pi }}{6}} \right\}\\b)\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\ \Rightarrow x + \dfrac{\pi }{4} = k\pi \\ \Rightarrow x = \dfrac{{ – \pi }}{4} + k\pi \\Do: – \dfrac{{5\pi }}{2} < x < 3\pi \\ \Rightarrow x \in \left\{ {\dfrac{{ – 9\pi }}{4};\dfrac{{ – 5\pi }}{4};\dfrac{{ – \pi }}{4};\dfrac{{3\pi }}{4};\dfrac{{7\pi }}{4};\dfrac{{11\pi }}{4}} \right\}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\sin x = \dfrac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
Do: – \dfrac{{5\pi }}{2} < x < 3\pi \\
\Rightarrow x \in \left\{ {\dfrac{{ – 11\pi }}{6};\dfrac{\pi }{6};\dfrac{{13\pi }}{6};\dfrac{{ – 7\pi }}{6};\dfrac{{5\pi }}{6};\dfrac{{17\pi }}{6}} \right\}\\
b)\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\Rightarrow x + \dfrac{\pi }{4} = k\pi \\
\Rightarrow x = \dfrac{{ – \pi }}{4} + k\pi \\
Do: – \dfrac{{5\pi }}{2} < x < 3\pi \\
\Rightarrow x \in \left\{ {\dfrac{{ – 9\pi }}{4};\dfrac{{ – 5\pi }}{4};\dfrac{{ – \pi }}{4};\dfrac{{3\pi }}{4};\dfrac{{7\pi }}{4};\dfrac{{11\pi }}{4}} \right\}
\end{array}$